20 % of a first order reaction was found to be completed at 10: 00 a . m at 11.30 a . m . on the same day, 20 % of the reaction was found to be remaining. The half life period in minutes of the reaction is

Step-by-step solution

For t=90 ~min (from 10 a.m. to 11: 30 a.m.) As 20 % was already completed a=80 aligned & (a-x)=(100-80)=20 & k= 2.303 t a a-x & k= 2.303 90 80 20 & k= 2.303 90 4= 2.303 0.6020 90 & k=0.015 ~min aligned aligned & t_ 1 / 2 = 0.693 k & t_ 1 / 2 = 0.693 0.015 =46.2 45 ~min aligned